First note since $
X \succeq xx^T \Leftrightarrow \begin{bmatrix}
X & x \\
x^T & 1
\end{bmatrix} \succeq 0
$, it makes sense to look at the problem as one in $S^{n+1}$.
Let $P_1, P_2$ be linear projections from $S^{n+1}$ to $S^{n} \times R^n$ and $R$ respectively:
$$
P_1 \left( \begin{bmatrix}
X & x \\
x^T & y
\end{bmatrix} \right) := (X,x), \quad
P_2 \left( \begin{bmatrix}
X & x \\
x^T & y
\end{bmatrix} \right) := y
$$
So then define the subsets of $S^{n+1}$:
$$
\begin{aligned}
D_1 &:= \{ W \in S^{n+1} : W \succeq 0, ~ tr(W) \le 2\}\\
D_2 &:= \{ W \in S^{n+1} : P_2(W) = 1\}
\end{aligned}
$$
Then the intersection of these is equivalent to $C$ in the following sense:
$$
(X,x) \in C \Leftrightarrow P_1^T(X,x) + P_2^T(1) \in D_1 \cap D_2
$$
Note that it is possible to project onto $D_1$ using eigenvalue decomposition and projection onto $D_2$ is trivial, since it's just a plane. So we can almost get a closed form expression for the projection onto the intersection.
Using the notation $
\Pi_A (z) = \mathrm{argmin}_{x \in A} \|x-z\|^2
$ for nonlinear projection onto a set, we have:
$$
\Pi_C (Z,z) = P_1\left( \Pi_{D_1} \left( \begin{bmatrix}
Z & z \\
z^T & y^*
\end{bmatrix} \right) \right)
$$
Where $y^*$ is the value satisfies the following:
$$
1 = P_2\left( \Pi_{D_1} \left( \begin{bmatrix}
Z & z \\
z^T & y^*
\end{bmatrix} \right) \right)
$$
This can be solved numerically via a 1-dimensional root-finding method. I'm not sure if it's possible to avoid calculating a full eigenvalue decomposition every time you project onto to $D_1$.
